Difference between revisions of "1953 AHSME Problems/Problem 1"

(Created page with "The boy buys <math>3</math> oranges for <math>10</math> cents or <math>1</math> orange for <math>\frac{10}{3}</math> cents. He sells them at <math>\frac{20}{5}=4</math> cents...")
 
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The boy buys <math>3</math> oranges for <math>10</math> cents or <math>1</math> orange for <math>\frac{10}{3}</math> cents.  He sells them at <math>\frac{20}{5}=4</math> cents each.
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A boy buys oranges at <math>3</math> for <math>10</math> cents. He will sell them at <math>5</math> for <math>20</math> cents. In order to make a profit of <math> </math>1.00<math>, he must sell:
That means for every orange he sells, he makes a profit of <math>4-\frac{10}{3}=\frac{2}{3}</math> cents.
 
  
To make a profit of <math>100</math> cents, he needs to sell <math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}</math>
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</math>\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\  \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}<math>
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==Solution==
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The boy buys </math>3<math> oranges for </math>10<math> cents or </math>1<math> orange for </math>\frac{10}{3}<math> cents.  He sells them at </math>\frac{20}{5}=4<math> cents each. 
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That means for every orange he sells, he makes a profit of </math>4-\frac{10}{3}=\frac{2}{3}<math> cents. 
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To make a profit of </math>100<math> cents, he needs to sell </math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$
  
 
~mathsolver101
 
~mathsolver101

Revision as of 13:11, 31 July 2015

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$ (Error compiling LaTeX. Unknown error_msg)1.00$, he must sell:$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$==Solution==

The boy buys$ (Error compiling LaTeX. Unknown error_msg)3$oranges for$10$cents or$1$orange for$\frac{10}{3}$cents.  He sells them at$\frac{20}{5}=4$cents each.   That means for every orange he sells, he makes a profit of$4-\frac{10}{3}=\frac{2}{3}$cents.

To make a profit of$ (Error compiling LaTeX. Unknown error_msg)100$cents, he needs to sell$\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$

~mathsolver101